Solvent selection — water, supercritical CO₂, ionic liquids: Exercises

Before we begin, one symbol we will reuse — introduced from zero so no line surprises you.

Look at the map below — it is the single picture the whole scCO₂ story lives on.

Figure — Solvent selection — water, supercritical CO₂, ionic liquids

The red dot is the critical point. The red shaded corner (top-right) is the supercritical region: you must be to the right of the red dot (hotter than $T_{c}$ ) and above it (higher pressure than $P_{c}$ ) at the same time. See Phase Diagrams & Critical Point for the full geometry.

Level 1 — Recognition

L1.1 Name the three "greener" solvents discussed in this topic, and for each give the ONE problem it primarily fixes (toxicity, recovery, or volatility).

L1.2 True or false: "Supercritical CO₂ means the CO₂ is extremely hot." Correct the statement if false.

L1.3 A solvent has a vapour pressure of essentially zero at room temperature and is a salt that is liquid below 100 °C. Which of the three solvents is it?

Recall Solution — L1

L1.1

Water → fixes toxicity/flammability (it is non-toxic, non-flammable, cheap).
Supercritical CO₂ → fixes recovery (depressurize and it flashes off as gas, zero residue).
Ionic liquid → fixes volatility (negligible vapour pressure ⇒ no VOC emission; see VOCs and Air Pollution).

L1.2 False. "Supercritical" means above the critical point in both $T$ and $P$ . For CO₂ that corner sits at only $31.1 ° C$ — near room temperature. It is about phase behaviour, not extreme heat.

L1.3 An ionic liquid. The two clues — a salt liquid below 100 °C and near-zero vapour pressure — are its defining traits.

Level 2 — Application

L2.1 CO₂ is held at $T = 40 ° C$ and $P = 90 bar$ . Is it supercritical? Show the two-part test.

L2.2 CO₂ is held at $T = 40 ° C$ and $P = 60 bar$ . Supercritical or not? Which condition fails?

L2.3 Convert the critical point of CO₂ to kelvin and MPa, then confirm it matches the given values $304.25 K$ and $7.38 MPa$ .

Recall Solution — L2

The test (from the definition): supercritical requires $T > T_{c}$ AND $P > P_{c}$ , with $T_{c} = 31.1 ° C$ , $P_{c} = 73.8 bar$ .

L2.1 Check both:

$T$ : $40 ° C > 31.1 ° C$ ✓
$P$ : $90 bar > 73.8 bar$ ✓

Both pass ⇒ supercritical. On the map (figure s01) this point sits inside the red corner.

L2.2 Check both:

$T$ : $40 ° C > 31.1 ° C$ ✓
$P$ : $60 bar > 73.8 bar$ ? No, $60 < 73.8$ ✗

The pressure condition fails ⇒ not supercritical. It is hot enough but not squeezed enough; it is a plain gas here.

L2.3 $T_{c} = 31.1 ° C + 273.15 = 304.25 K ✓$ $P_{c} = 73.8 bar \times \frac{1 MPa}{10 bar} = 7.38 MPa ✓$

Level 3 — Analysis

L3.1 Near the critical point, a modest pressure increase produces a huge change in solvent power. Using the compressibility relation, explain why this happens and why it is useful. $κ_{T} = - \frac{1}{V} (\frac{\partial V}{\partial P})_{T} = \frac{1}{ρ} (\frac{\partial ρ}{\partial P})_{T}$

L3.2 A Diels–Alder reaction between two non-polar molecules runs faster as a suspension in water than in an organic solvent, even though the reactants barely dissolve. Explain the two reasons (concentration and transition state), naming the effect. Link the relevant concept.

L3.3 Study the density-vs-pressure sketch below (figure s02). Two operating points, $A$ and $B$ , differ by the same $Δ P$ . Explain why the pressure "dial" is far more powerful near the steep red section than on the flat black section.

Recall Solution — L3

L3.1 Read the relation right to left. The term $(\partial ρ / \partial P)_{T}$ means "how much does density change when I nudge pressure." The relation ties it to $κ_{T}$ , the isothermal compressibility (how squishable the fluid is at fixed $T$ ).

At the critical point the pressure–volume curve is momentarily flat: $(\partial P / \partial V)_{T} = 0$ .
Its reciprocal, $(\partial V / \partial P)_{T}$ , therefore blows up ⇒ $κ_{T} \to \infty$ .
A near-infinite compressibility means a small $Δ P$ produces a large $Δ ρ$ .

Why useful: solvent power tracks density. So a tiny turn of the pressure knob near $T_{c}$ swings the dissolving strength widely — a continuous "solvent-strength dial." That is the engineering heart of scCO₂.

L3.2 See Hydrophobic Effect.

Concentration: water refuses to mingle with the non-polar reactants, forcing them to huddle together at the interface. Their effective local concentration rockets, and reaction rate rises with concentration.
Transition state: the Diels–Alder transition state is more compact and slightly more polar than the loose reactants. Water stabilises this polar TS, lowering the activation energy $E_{a}$ , which speeds the reaction.

Net: greener (no organic solvent) and faster.

L3.3 On the flat black stretch (far from $T_{c}$ ) the fluid is barely compressible: the same $Δ P$ barely moves $ρ$ , so solvent power hardly changes. On the steep red stretch (near $T_{c}$ ) the fluid is extremely compressible ( $κ_{T}$ large), so the identical $Δ P$ causes a big $Δ ρ$ and a big change in dissolving power. Same knob, wildly different sensitivity — operate where the curve is steep.

Level 4 — Synthesis

L4.1 You must decaffeinate coffee for the food market. Design the solvent process step by step, justifying each choice: solvent, operating $(T, P)$ region, a co-solvent tweak, and separation.

L4.2 You have an expensive transition-metal catalyst you want to reuse across many batches. Design a biphasic system using an ionic liquid. Explain what goes in each layer, why they stay separate, and why the IL specifically avoids contaminating the product. Link the recovery concept.

L4.3 A reagent in your reaction is a Grignard reagent (reacts violently with any O–H bond). A colleague says "just use water, it's the greenest solvent." Give the correct decision and the specific chemical reason water fails here.

Recall Solution — L4

L4.1 Decaffeination process

Choose scCO₂. Why: food-grade, non-toxic, non-flammable, and leaves zero residue on depressurising — essential for food.
Operate just above $T_{c}, P_{c}$ (≈ $31 ° C$ , ≈ $74 bar$ ). Why: the mild temperature protects delicate flavour compounds while still being supercritical.
Add a little water as co-solvent. Why: caffeine is somewhat polar; pure CO₂ is non-polar, so a splash of water nudges the solvent polarity up and lifts the caffeine out.
Depressurise. Why: CO₂ flashes to gas, dropping the dissolved caffeine cleanly — perfect separation, clean beans, recyclable CO₂ (closed loop).

L4.2 Biphasic IL catalysis — see Catalysis & Catalyst Recovery.

IL layer: dissolve the catalyst here; ILs are polar and hold ionic/metal species well.
Second layer: an immiscible organic (or scCO₂) layer where the product collects.
Why they stay separate: the two liquids are immiscible, so they form two phases like oil and water — pour off the product layer, keep the catalyst locked in the IL for the next batch.
Why IL specifically: its negligible vapour pressure means the IL never evaporates into the product stream, so the product is never contaminated by solvent.

L4.3 Grignard case

Correct decision: do not use water. Use an inert dry solvent (this is a "right tool, right job" call).
Reason: a Grignard reagent reacts instantly and violently with any O–H bond. Water is loaded with O–H bonds, so it would hydrolyse and destroy the reagent before it could react as intended. Water's safety credentials are irrelevant if it eats your reactant. Greenest-on-paper ≠ compatible.

Level 5 — Mastery

L5.1 A start-up markets a new ionic liquid as "100% green because it never evaporates." Critique this using a whole-life-cycle argument. Name at least three impact categories the claim ignores and the framework that captures them.

L5.2 An environmental group protests that "scCO₂ processes pump more CO₂ into the atmosphere and worsen global warming." Rebut this with the correct process fact.

L5.3 Rank a decision. A reaction can run in (a) water, (b) scCO₂, or (c) an ionic liquid. The product must be totally solvent-free, the substrate is non-polar, and the budget is tight but not zero. Which solvent, and give the two strongest reasons plus the one drawback you accept?

Recall Solution — L5

L5.1 Zero volatility only removes air pollution — it says nothing about the rest of the life cycle. See Life Cycle Assessment (LCA) and Principles of Green Chemistry. Ignored categories:

Aquatic toxicity — many ILs are toxic to water organisms.
Biodegradability — many ILs persist in soil and water.
Synthesis burden — ILs are energy-intensive and often multi-step to manufacture (poor atom economy).

Framework: greenness must be judged over the whole life cycle, from cradle to grave, not on a single headline virtue.

L5.2 The CO₂ used in scCO₂ processes is captured and recycled in a closed loop — frequently a by-product recovered from other industrial streams. It is contained, compressed, used, depressurised, and re-compressed for reuse; it is not newly generated and not vented. So the process does not add net CO₂ to the atmosphere.

L5.3 Choose (b) scCO₂.

Reason 1 — solvent-free product: depressurising flashes the CO₂ off entirely, leaving zero residue — the hard requirement.
Reason 2 — dissolves non-polar substrate: scCO₂ is itself non-polar, so it dissolves the non-polar substrate well (water would not; an IL is polar and costly).
Accepted drawback: you must buy and run high-pressure equipment (≥ 74 bar). Given the budget is "tight but not zero," this is tolerable, and it beats water (won't dissolve the substrate, hard to remove) and an IL (expensive, and would need extra separation to guarantee a solvent-free product).